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Question #25252
Show that from: rad (R[t]) = (Nil*R)[t] follows Kothe’s Conjecture. (“The sum of two nil left ideals in any ring is nil”.)
We shall prove Kothe&rsquo;s Conjecture inthe alternative form: If R is a ring such that Nil*R = 0, thenevery nil left ideal I &sube; R is zero. Itsuffices to show that, for any f(t) &isin; I[t], 1 + f(t) has a leftinverse in R[t], for then I[t] &sube; rad R[t], and rad R[t] = 0 by assumption.Consider the ring R&#039; = I &oplus; Z, obtained by formally adjoining anidentity 1&#039; = (0, 1) to the &ldquo;rng&rdquo; I. It is easy to seethat Nil*R&#039; = I, so by assumption, rad R&#039;[t] = I[t].Given f(t) &isin; I[t],1&#039; + f(t) has then a left inverse, say 1&#039; + g(t),in R&#039;[t]. From1&#039; = (1&#039; + g(t))(1&#039; + f(t)),we have
&minus;g(t) = f(t) + g(t)f(t)&isin; I[t],
since I[t] is an idealin R&#039;[t]. Now the equation 1 = (1+g(t))(1 + f(t))implies that 1 + f(t) has a left inverse 1 + g(t) &isin; R[t], as desired.

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